Under conditions \({(\ref {cond 1})}\) , we have \(h_2(p)=h_2(q)=h_2(-q)=h_2(2q)=h_2(-1)=h_2(2)=h_2(-2)=1\) ,
\(h_2(-p)=h_2(2p)=h_2(-2p)=h_2(pq)=h_2(-pq)=h_2(2pq)=2\)  and \(h_2(-2pq)=4\)  (cf. [1]}, [2]}). So by Wada's class number formula (cf. [3]}) one gets
\(\begin{array}{ll}h_2(\mathbb {K})&=\frac{1}{2^{16}}q(\mathbb {K}) h_2(-1) h_2(2) h_2(-2) h_2(p) h_2(-p) h_2(q)\\&\qquad h_2(-q) h_2(2p)h_2(-2p) h_2(2q)h_2(-2q) h_2(pq) h_2(-pq) h_2(2pq) h_2(-2pq)\\&=\frac{1}{2^{16}}\cdot q(\mathbb {K})\cdot 1\cdot 1 \cdot 1 \cdot 1 \cdot 2 \cdot 1 \cdot 1 \cdot 2 \cdot 2\cdot 1\cdot h_2(-2q)\cdot 2\cdot 2\cdot 2\cdot 4, \\&= \frac{1}{2^{8}}.q(\mathbb {K}).h_2(-2q).\end{array}\) 
On the other hand, by Theorem REF  we have \(h_2(\mathbb {K})=h_2(-2q)\) . So obviously we must have \(q(\mathbb {K})= 2^{8}\) .

Suppose that the unit group of \(\mathbb {K}^+ \)  takes the form in the first item of Lemma REF , then a FSU of \(\mathbb {K}^+ \)  is
\( \lbrace  \varepsilon _{2}, \varepsilon _{p}, \sqrt{\varepsilon _{q}}, \sqrt{\varepsilon _{2q}}, \sqrt{\varepsilon _{pq}}, \sqrt{\varepsilon _{2pq}}, \sqrt{\varepsilon _{2}\varepsilon _{p}\varepsilon _{2p}} \rbrace  =\lbrace \alpha _1, \alpha _2, ..., \alpha _7 \rbrace .\) 
Thus by Lemma REF  a FSU of \(\mathbb {K}\)  is \(\lbrace \alpha _1, \alpha _2, ..., \alpha _7 \rbrace \)  or \(\lbrace  \alpha _{i_1}, ..., \alpha _{i_5}, \alpha _{i_0}, \sqrt{\zeta _8\alpha } \rbrace \)  with \(i_k\in \lbrace 1, ..., 7\rbrace \)  and
\(\alpha =\alpha _1^{r_1}\alpha _2^{r_2}\cdots \alpha _7^{r_7}\) , where \(r_k\in \lbrace 0, 1\rbrace  \) , and \(\alpha _{i_0}\in \lbrace \varepsilon _{2}, \varepsilon _{p}\rbrace \) , for some \(i_0\) . Thus \(q(\mathbb {K})\le 2^{7}\) . Which is absurd.

Assume now that the unit group of \(\mathbb {K}\)  takes the form in the first item of Lemma REF , then \(q(\mathbb {K})\le 2^{7}\) . which is also absurd.

Thus the only possible case is the one which is given by the second item of Lemma REF . Which completes the proof.
